Answer 1

0.003MHCl

`H_2O + HCl ↔  H_3O^+   + Cl^(-)` 

Since HCl is completely ionized,

`[H_3O^+] = [HCl]`

`=>[H_3O^+] = 0.003`

Now

`pH = - log[H_3O^+] = -log(.003)`

= 2.52

Hence, the pH of the solution is 2.52.